Question

a,b,c,d \(\in\) R^\(+\)

^{\(\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+d\right)\left(b+c\right)}\)}

Answer 1

BĐT \(\Leftrightarrow\left(\sqrt{ab}+\sqrt{cd}\right)^2\le\left(a+d\right)\left(b+c\right)\)

\(\Leftrightarrow ab+2\sqrt{abcd}+cd\le ab+ac+bd+dc\)

\(\Leftrightarrow2\sqrt{abcd}\le ac+bd\)

\(\Leftrightarrow0\le\left(\sqrt{ac}-\sqrt{bd}\right)^2\) ( luôn đúng )

Dấu "=" xảy ra khi \(\sqrt{ac}=\sqrt{bd}\Leftrightarrow ac=bd\Leftrightarrow\frac{a}{b}=\frac{d}{c}\)

Answer 2

bunhiacopxki thử xem nha!

\(VP=\sqrt{\left(\sqrt{a}^2+\sqrt{d}^2\right)\left(\sqrt{b}^2+\sqrt{c}^2\right)}\)

\(\ge\sqrt{\left(\sqrt{ad}+\sqrt{cd}\right)^2}=\sqrt{ad}+\sqrt{cd}=VT\)

ta có dpcm.